[tex] x^{2} - 3x - 18 = 0[/tex] factors → [tex](x - 6) * (x+ 3)[/tex]
If you want to find your roots, use Bhaskara's formula
[tex] x^{2} - 3x - 18 = 0[/tex]
a = 1; b = - 3, c = - 18
[tex]\Delta = b^2 - 4*a*c[/tex]
[tex]\Delta = (-3)^2 - 4*1*(-18)[/tex]
[tex]\Delta = 9 + 72[/tex]
[tex]\Delta = 81[/tex]
[tex]x = \frac{-b\pm \sqrt{\Delta} }{2*a} [/tex]
[tex]x = \frac{-(-3)\pm \sqrt{81} }{2*(-3)} [/tex]
[tex]x = \frac{3\pm 9 }{-6}[/tex]
[tex]x' = \frac{3-9}{-6} = \frac{-6}{-6} \rightarrow \boxed{x' = 1}[/tex]
[tex]x'' = \frac{3+9}{-6} = \frac{12}{-6} \rightarrow \boxed{x'' = -2}[/tex]
But, the answer to your question are the factors of the equation, here is:
ANSWER:
[tex](x - 6) * (x+ 3)[/tex]
x - 6 and x + 3
